Chapter 16, Problem 44P

(a)

To determine

The electric flux through the circle when its face is perpendicular to the filed lines.

Answer to Problem 44P

The required direction is right.

Explanation of Solution

Given info:

The radius of the circle is $r=18\text{cm}$ .

The magnitude of electric field is $E=5.8\times {10}^2\text{N/C}$ .

When the face is perpendicular to the electric field $\theta =0°$ .

Formula used:

The area of the circle is,

  $A=\pi r^2$

The electric flux through the circle is,

  ${\varphi }_E=EA\cos \theta$

Calculation:

Substituting the given values in formula $A=\pi r^2$ , we get

  $\begin{array}{c}A=\pi {\left(0.18\right)}^2\\ =0.102{\text{m}}^{\text{2}}\end{array}$

Substituting the given values in formula ${\varphi }_E=EA\cos \theta$ , we get

  $\begin{array}{c}{\varphi }_E=\left(5.8\times {10}^2\text{N/C}\right)\left(0.102{\text{m}}^{\text{2}}\right)\cos 0°\\ =59.16{\text{Nm}}^{\text{2}}\text{/C}\end{array}$

Conclusion:

Thus, the electric flux through the circle when its face is perpendicular to the filed lines is $59.16{\text{Nm}}^{\text{2}}\text{/C}$ .

(b)

To determine

The electric flux through the circle when its face is at $45°$ to the filed lines.

Answer to Problem 44P

The electric flux through the circle when its face is at $45°$ to the filed lines is $41.8{\text{Nm}}^{\text{2}}\text{/C}$ .

Explanation of Solution

Given info:

The radius of the circle is $r=18\text{cm}$ .

The magnitude of electric field is $E=5.8\times {10}^2\text{N/C}$ .

When the face is at $45°$ to the electric field $\theta =45°$ .

Formula used:

The electric flux through the circle is,

  ${\varphi }_E=EA\cos \theta$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}{\varphi }_E=\left(5.8\times {10}^2\text{N/C}\right)\left(0.102{\text{m}}^{\text{2}}\right)\cos 45°\\ =41.8{\text{Nm}}^{\text{2}}\text{/C}\end{array}$

Conclusion:

Thus, the electric flux through the circle when its face is at $45°$ to the filed lines is $41.8{\text{Nm}}^{\text{2}}\text{/C}$ .

(c)

To determine

The electric flux through the circle when its face is parallel to the filed lines.

Answer to Problem 44P

The electric flux through the circle when its face is parallel to the filed lines is $0{\text{Nm}}^{\text{2}}\text{/C}$ .

Explanation of Solution

Given info:

The radius of the circle is $r=18\text{cm}$ .

The magnitude of electric field is $E=5.8\times {10}^2\text{N/C}$ .

When the face is parallel to the electric field $\theta =90°$ .

Formula used:

The electric flux through the circle is,

  ${\varphi }_E=EA\cos \theta$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}{\varphi }_E=\left(5.8\times {10}^2\text{N/C}\right)\left(0.102{\text{m}}^{\text{2}}\right)\cos 90°\\ =0{\text{Nm}}^{\text{2}}\text{/C}\end{array}$

Conclusion:

Thus, the electric flux through the circle when its face is parallel to the filed lines is $0{\text{Nm}}^{\text{2}}\text{/C}$ .